AllLife Bank Customer Segmentation¶

Context¶

AllLife Bank wants to focus on its credit card customer base in the next financial year. They have been advised by their marketing research team, that the penetration in the market can be improved. Based on this input, the Marketing team proposes to run personalized campaigns to target new customers as well as upsell to existing customers. Another insight from the market research was that the customers perceive the support services of the back poorly. Based on this, the Operations team wants to upgrade the service delivery model, to ensure that customer queries are resolved faster. Head of Marketing and Head of Delivery both decide to reach out to the Data Science team for help

Objective¶

To identify different segments in the existing customer, based on their spending patterns as well as past interaction with the bank, using clustering algorithms, and provide recommendations to the bank on how to better market to and service these customers.

Key Questions¶

  • Exploratory Data Analysis to understand the data better
  • Data preprocessing to prepare the data for modeling
  • KMeans and Hierarchical clustering cluster analysis and profiling
  • Comparing K-means vs Hierarchical Clustering
  • Actional insights and recommendation to bank

Data Description¶

The data provided is of various customers of a bank and their financial attributes like credit limit, the total number of credit cards the customer has, and different channels through which customers have contacted the bank for any queries (including visiting the bank, online and through a call center).

Data Dictionary

  • Sl_No: Primary key of the records
  • Customer Key: Customer identification number
  • Average Credit Limit: Average credit limit of each customer for all credit cards
  • Total credit cards: Total number of credit cards possessed by the customer
  • Total visits bank: Total number of visits that customer made (yearly) personally to the bank
  • Total visits online: Total number of visits or online logins made by the customer (yearly)
  • Total calls made: Total number of calls made by the customer to the bank or its customer service department (yearly)

Importing necessary libraries¶

In [1]:
# Libraries to help with reading and manipulating data
import numpy as np
import pandas as pd

# Libraries to help with data visualization
import matplotlib.pyplot as plt
import seaborn as sns

# Removes the limit for the number of displayed columns
pd.set_option("display.max_columns", None)
# Sets the limit for the number of displayed rows
pd.set_option("display.max_rows", 200)

# to scale the data using z-score
from sklearn.preprocessing import StandardScaler

# to compute distances
from scipy.spatial.distance import pdist, cdist

# to perform k-means clustering and compute silhouette scores
from sklearn.cluster import KMeans
from sklearn.metrics import silhouette_score

# to visualize the elbow curve and silhouette scores
from yellowbrick.cluster import KElbowVisualizer, SilhouetteVisualizer

# to perform hierarchical clustering, compute cophenetic correlation, and create dendrograms
from sklearn.cluster import AgglomerativeClustering
from scipy.cluster.hierarchy import dendrogram, linkage, cophenet

# to perform PCA
from sklearn.decomposition import PCA

import warnings
warnings.filterwarnings('ignore')

#Format:
pd.options.display.float_format = '{:.4f}'.format

%matplotlib inline

Read the dataset¶

In [2]:
# loading the dataset
data = pd.read_excel("Credit_Card_Customer_Data.xlsx")

View the random 10 rows of the dataset.¶

In [3]:
# viewing a random sample of the dataset
data.sample(n=10, random_state=1)
Out[3]:
Sl_No Customer Key Avg_Credit_Limit Total_Credit_Cards Total_visits_bank Total_visits_online Total_calls_made
547 548 38125 26000 4 5 2 4
353 354 94437 9000 5 4 1 3
499 500 65825 68000 6 4 2 2
173 174 38410 9000 2 1 5 8
241 242 81878 10000 4 5 1 3
341 342 70779 18000 4 3 2 0
647 648 79953 183000 9 0 9 2
218 219 28208 19000 3 1 5 7
120 121 16577 10000 4 2 4 6
134 135 31256 13000 4 1 5 7

Dataset Summary¶

Shape of the data¶

In [4]:
print('There are {row} records, and {col} columns in the dataset'.format(row=data.shape[0], col=data.shape[1]))

There are 660 records, and 7 columns in the dataset
  • The dataset has 660 rows and 7 columns
In [5]:
# copying the data to another variable to avoid any changes to original data
df = data.copy()

Check the data types of the columns for the dataset.¶

In [6]:
# checking datatypes and number of non-null values for each column
df.info()

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 660 entries, 0 to 659
Data columns (total 7 columns):
 #   Column               Non-Null Count  Dtype
---  ------               --------------  -----
 0   Sl_No                660 non-null    int64
 1   Customer Key         660 non-null    int64
 2   Avg_Credit_Limit     660 non-null    int64
 3   Total_Credit_Cards   660 non-null    int64
 4   Total_visits_bank    660 non-null    int64
 5   Total_visits_online  660 non-null    int64
 6   Total_calls_made     660 non-null    int64
dtypes: int64(7)
memory usage: 36.2 KB

Observations

  • All the variables are numeric variables.
  • There are no missing values present in any of the variables.

Summary of the dataset.¶

In [7]:
# Let's look at the statistical summary of the data
df.describe(include="all").T
Out[7]:
count mean std min 25% 50% 75% max
Sl_No 660.0000 330.5000 190.6699 1.0000 165.7500 330.5000 495.2500 660.0000
Customer Key 660.0000 55141.4439 25627.7722 11265.0000 33825.2500 53874.5000 77202.5000 99843.0000
Avg_Credit_Limit 660.0000 34574.2424 37625.4878 3000.0000 10000.0000 18000.0000 48000.0000 200000.0000
Total_Credit_Cards 660.0000 4.7061 2.1678 1.0000 3.0000 5.0000 6.0000 10.0000
Total_visits_bank 660.0000 2.4030 1.6318 0.0000 1.0000 2.0000 4.0000 5.0000
Total_visits_online 660.0000 2.6061 2.9357 0.0000 1.0000 2.0000 4.0000 15.0000
Total_calls_made 660.0000 3.5833 2.8653 0.0000 1.0000 3.0000 5.0000 10.0000

Observations

  • There are 660 rows and no missing values in any feature.
  • Sl.No and Customer_Key feature will anyway be removed as it won't help much in analysis
  • Credit limit average is around 35K with 50% of customers having a credit limit less than 18K. So, this feature is heavily right skewed. Also the Standard deviation is 37K which is also huge.
  • Credit Card count ranges from 1 to 10 cards with an avg of 5 and also 75% of customers have upto 6 only.
  • On average, most customer interactions are through calls, then online. Also, some customers never contacted/visited the bank.

Data PreProcessing¶

In [8]:
# dropping the serial no. column as it does not provide any information
df.drop(["Sl_No"], axis=1, inplace=True)
In [9]:
# checking for missing values
df.isna().sum()
Out[9]:
Customer Key           0
Avg_Credit_Limit       0
Total_Credit_Cards     0
Total_visits_bank      0
Total_visits_online    0
Total_calls_made       0
dtype: int64
In [10]:
df.duplicated().sum()
Out[10]:
0
In [11]:
#Checking for number of unique values in each variables
df.nunique()
Out[11]:
Customer Key           655
Avg_Credit_Limit       110
Total_Credit_Cards      10
Total_visits_bank        6
Total_visits_online     16
Total_calls_made        11
dtype: int64
  • It looks like there are 5 duplicate customer key values present in the dataframe. Lets analyze the records
In [12]:
data[data['Customer Key'].isin(data[data['Customer Key'].duplicated()]['Customer Key'].tolist())].sort_values('Customer Key')
Out[12]:
Sl_No Customer Key Avg_Credit_Limit Total_Credit_Cards Total_visits_bank Total_visits_online Total_calls_made
48 49 37252 6000 4 0 2 8
432 433 37252 59000 6 2 1 2
4 5 47437 100000 6 0 12 3
332 333 47437 17000 7 3 1 0
411 412 50706 44000 4 5 0 2
541 542 50706 60000 7 5 2 2
391 392 96929 13000 4 5 0 0
398 399 96929 67000 6 2 2 2
104 105 97935 17000 2 1 2 10
632 633 97935 187000 7 1 7 0
  • Its not necessary to remove the Customer Key duplicate values as its wont impact much
  • Also this key column will be removed for further classification analysis
  • If at all there are some duplicate records/rows persent in the data, it wont impact much as we are going to cluster the records.
In [13]:
# Subseting the data for further analysis
subset_df = df.drop("Customer Key", axis=1)
In [14]:
subset_df.columns
Out[14]:
Index(['Avg_Credit_Limit', 'Total_Credit_Cards', 'Total_visits_bank',
       'Total_visits_online', 'Total_calls_made'],
      dtype='object')
In [15]:
num_cols = subset_df.select_dtypes(include=np.number).columns.tolist()
num_cols
Out[15]:
['Avg_Credit_Limit',
 'Total_Credit_Cards',
 'Total_visits_bank',
 'Total_visits_online',
 'Total_calls_made']
In [16]:
value_count_cols = ['Total_Credit_Cards',
 'Total_visits_bank',
 'Total_visits_online',
 'Total_calls_made']
for column in value_count_cols:
    print(df[column].value_counts())
    print("-" * 50)

4     151
6     117
7     101
5      74
2      64
1      59
3      53
10     19
9      11
8      11
Name: Total_Credit_Cards, dtype: int64
--------------------------------------------------
2    158
1    112
0    100
3    100
5     98
4     92
Name: Total_visits_bank, dtype: int64
--------------------------------------------------
2     189
0     144
1     109
4      69
5      54
3      44
15     10
7       7
10      6
12      6
8       6
11      5
13      5
9       4
14      1
6       1
Name: Total_visits_online, dtype: int64
--------------------------------------------------
4     108
0      97
2      91
1      90
3      83
6      39
7      35
9      32
8      30
5      29
10     26
Name: Total_calls_made, dtype: int64
--------------------------------------------------

Observation

  • Total Credit Card: 151 customers have 4 credit cards and only 19 have 10 credit cards
  • Total Bank Visits: Max of 158 customers visited bank twice and 100 customers never visited bank with 112 visited atleast once
  • Total Online Visits: 144 customers never visited, 109 visited only once and 189 visited twice.
  • Total Calls made: 97 customers never called bank and max of 108 customers called 4 times.

EDA¶

Univariate Analysis¶

In [17]:
# function to plot a boxplot and a histogram along the same scale.


def histogram_boxplot(data, feature, figsize=(12, 7), kde=False, bins=None):
    """
    Boxplot and histogram combined

    data: dataframe
    feature: dataframe column
    figsize: size of figure (default (12,7))
    kde: whether to the show density curve (default False)
    bins: number of bins for histogram (default None)
    """
    f2, (ax_box2, ax_hist2) = plt.subplots(
        nrows=2,  # Number of rows of the subplot grid= 2
        sharex=True,  # x-axis will be shared among all subplots
        gridspec_kw={"height_ratios": (0.25, 0.75)},
        figsize=figsize,
    )  # creating the 2 subplots
    sns.boxplot(
        data=data, x=feature, ax=ax_box2, showmeans=True, color="violet"
    )  # boxplot will be created and a star will indicate the mean value of the column
    sns.histplot(
        data=data, x=feature, kde=kde, ax=ax_hist2, bins=bins, palette="winter"
    ) if bins else sns.histplot(
        data=data, x=feature, kde=kde, ax=ax_hist2
    )  # For histogram
    ax_hist2.axvline(
        data[feature].mean(), color="green", linestyle="--"
    )  # Add mean to the histogram
    ax_hist2.axvline(
        data[feature].median(), color="black", linestyle="-"
    )  # Add median to the histogram
In [18]:
# selecting numerical columns
for item in num_cols:
    histogram_boxplot(subset_df, item, bins=50, kde=True, figsize=(10, 5))
In [19]:
# function to create labeled barplots


def labeled_barplot(data, feature, perc=False, n=None):
    """
    Barplot with percentage at the top

    data: dataframe
    feature: dataframe column
    perc: whether to display percentages instead of count (default is False)
    n: displays the top n category levels (default is None, i.e., display all levels)
    """

    total = len(data[feature])  # length of the column
    count = data[feature].nunique()
    if n is None:
        plt.figure(figsize=(count + 2, 6))
    else:
        plt.figure(figsize=(n + 2, 6))

    plt.xticks(rotation=90, fontsize=15)
    ax = sns.countplot(
        data=data,
        x=feature,
        palette="Paired",
        order=data[feature].value_counts().index[:n].sort_values(),
    )

    for p in ax.patches:
        if perc == True:
            label = "{:.1f}%".format(
                100 * p.get_height() / total
            )  # percentage of each class of the category
        else:
            label = p.get_height()  # count of each level of the category

        x = p.get_x() + p.get_width() / 2  # width of the plot
        y = p.get_height()  # height of the plot

        ax.annotate(
            label,
            (x, y),
            ha="center",
            va="center",
            size=12,
            xytext=(0, 5),
            textcoords="offset points",
        )  # annotate the percentage

    plt.show()  # show the plot
In [20]:
for i in range(len(value_count_cols)):
    labeled_barplot(data, value_count_cols[i])
    labeled_barplot(data, value_count_cols[i], perc=True)
In [21]:
all_col = subset_df.iloc[:,:].columns.tolist()
fig, axes = plt.subplots(1, 5,  figsize=(20, 6))
fig.suptitle('Histogram of numerical variables', fontsize=20)
counter = 0
for ii in range(5):
    sns.countplot(ax=axes[ii],x=subset_df[all_col[counter]])
    counter = counter+1

fig.tight_layout(pad=4.0)
In [22]:
## Number of customers who used all three services 
subset_df[(subset_df['Total_calls_made']!=0) & (subset_df['Total_visits_bank']!=0) & (subset_df['Total_visits_online']!=0)].count()
Out[22]:
Avg_Credit_Limit       356
Total_Credit_Cards     356
Total_visits_bank      356
Total_visits_online    356
Total_calls_made       356
dtype: int64
In [23]:
## Number of customers who never used all three services 
subset_df[(subset_df['Total_calls_made']==0) & (subset_df['Total_visits_bank']==0) & (subset_df['Total_visits_online']==0)].count()
Out[23]:
Avg_Credit_Limit       0
Total_Credit_Cards     0
Total_visits_bank      0
Total_visits_online    0
Total_calls_made       0
dtype: int64
In [24]:
from pandas_profiling import ProfileReport

profile = ProfileReport(subset_df)
In [25]:
profile.to_notebook_iframe()

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Observation of Univariate Analysis¶

  • Many outliers in average credit limit. High credit customers are causing skewness.
  • Majority of customers have 4 credit cards and visited bank & online twice
  • Online visits are mostly between 1 and 4 with some outliers with more than 7 and above.
  • There are multiple peaks in credit cards, visits online, total calls.
  • Every customer have used either of the options to contact bank
  • 356 customers used all 3 banking options (calls / online / visited) which is more than half of our customers.

Advantages of Univariate Analysis

  • Helps to check the distribution/spread of the data primarily using histograms and box plots.

  • Also Numerical feature can be plot using different options like violin plot and cumulative density distribution plot.

  • Best way to understand data before starting the analysis or modeling.

Bivariate Analysis¶

In [26]:
sns.pairplot(subset_df, diag_kind='kde');
In [27]:
for i in range(len(value_count_cols)):
    sns.lineplot(subset_df['Total_Credit_Cards'],subset_df[value_count_cols[i]])

Let's check for correlations.

In [28]:
plt.figure(figsize=(15, 7))
sns.heatmap(
    subset_df[num_cols].corr(), annot=True, vmin=-1, vmax=1, fmt=".2f", cmap="Spectral"
)
plt.show()

Observations on Bivariate Analysis¶

  • Average Credit Limit has slightly positive correlation with Total Credit Cards
  • Total Number of Visits Online has slightly negative correlation with Total Calls Made
  • Total Credit Cards and Total Calls Made are negatively correlated
  • Total Visits Online is also negatively correlated with Total Visits to Bank
  • One intresting observation is calls made to bank is reducing as the number of credit cards increased
  • Customers who have less credit cards like 1, 2 or 3 are calling bank frequently.
  • Vists to bank is more for customers having credit cards between 4 to 7
  • Also Visiting online customers have more credit cards 8 to 10

Missing Values Treatment¶

In [29]:
subset_df.isnull().sum()
Out[29]:
Avg_Credit_Limit       0
Total_Credit_Cards     0
Total_visits_bank      0
Total_visits_online    0
Total_calls_made       0
dtype: int64
  • There are no missing values in any column to treat before modelling

Outliers Treatment¶

Let's find the percentage of outliers, in each column of the data, using IQR.

Treating outliers

We will cap/clip the minimum and maximum value of these columns to the lower and upper whisker value of the boxplot found using Q1 - 1.5*IQR and Q3 + 1.5*IQR, respectively.

Note: Generally, a value of 1.5 * IQR is taken to cap the values of outliers to upper and lower whiskers but any number (example 0.5, 2, 3, etc) other than 1.5 can be chosen. The value depends upon the business problem statement.

In [30]:
# Finding the 25th percentile and 75th percentile for the numerical columns.
Q1 = data[num_cols].quantile(0.25)
Q3 = data[num_cols].quantile(0.75)

IQR = Q3 - Q1  # Inter Quantile Range (75th percentile - 25th percentile)

lower_whisker = (
    Q1 - 1.5 * IQR
)  # Finding lower and upper bounds for all values. All values outside these bounds are outliers
upper_whisker = Q3 + 1.5 * IQR
In [31]:
# Percentage of outliers in each column
((subset_df[num_cols] < lower_whisker) | (subset_df[num_cols] > upper_whisker)).sum() / subset_df.shape[
    0
] * 100
Out[31]:
Avg_Credit_Limit      5.9091
Total_Credit_Cards    0.0000
Total_visits_bank     0.0000
Total_visits_online   5.6061
Total_calls_made      0.0000
dtype: float64
In [32]:
# Plotting Boxplots of all our features to get idea of distribution and outliers
plt.figure(figsize=(18,6))
sns.boxplot(x="value", y="variable", data=pd.melt(subset_df))
plt.title('Boxplots of all variables', size=15)
plt.show()
In [33]:
Q1 = subset_df["Avg_Credit_Limit"].quantile(0.25)  # 25th quantile
Q3 = subset_df["Avg_Credit_Limit"].quantile(0.75)  # 75th quantile
IQR = Q3 - Q1  # Inter Quantile Range (75th perentile - 25th percentile)
lower_whisker = Q1 - 1.5 * IQR
upper_whisker = Q3 + 1.5 * IQR
print(lower_whisker)
print(upper_whisker)

-47000.0
105000.0
In [34]:
subset_df.loc[subset_df["Avg_Credit_Limit"] > 105000.0].sort_values("Avg_Credit_Limit", ascending=False)
Out[34]:
Avg_Credit_Limit Total_Credit_Cards Total_visits_bank Total_visits_online Total_calls_made
630 200000 10 0 13 0
650 195000 10 1 12 2
631 195000 8 0 15 0
632 187000 7 1 7 0
646 186000 7 0 8 1
639 184000 7 1 15 2
647 183000 9 0 9 2
624 178000 7 0 11 0
622 176000 10 1 15 2
637 173000 9 1 11 0
658 172000 10 1 15 0
654 172000 10 1 9 1
645 171000 10 0 15 0
659 167000 9 0 12 2
621 166000 9 1 12 2
623 166000 10 0 7 0
614 163000 8 1 7 1
633 163000 7 1 10 1
619 158000 7 0 13 0
612 157000 9 1 14 1
652 156000 8 1 8 0
626 156000 9 1 10 2
629 155000 8 0 7 2
638 153000 8 1 7 1
627 146000 10 0 12 1
657 145000 8 1 9 1
641 144000 10 0 10 2
617 136000 8 0 13 0
651 132000 9 1 12 2
615 131000 9 1 10 1
644 127000 10 1 15 1
636 126000 10 1 8 0
640 123000 8 1 15 2
618 121000 7 0 13 2
635 114000 10 1 7 2
649 112000 10 1 8 1
648 111000 8 1 7 0
620 108000 10 0 15 1
634 106000 8 0 8 1
In [35]:
Q1 = subset_df["Total_visits_online"].quantile(0.25)  # 25th quantile
Q3 = subset_df["Total_visits_online"].quantile(0.75)  # 75th quantile
IQR = Q3 - Q1  # Inter Quantile Range (75th perentile - 25th percentile)
lower_whisker = Q1 - 1.5 * IQR
upper_whisker = Q3 + 1.5 * IQR
print(lower_whisker)
print(upper_whisker)

-3.5
8.5
In [36]:
subset_df.loc[subset_df["Total_visits_online"] > 8.5].sort_values("Total_visits_online", ascending=False)
Out[36]:
Avg_Credit_Limit Total_Credit_Cards Total_visits_bank Total_visits_online Total_calls_made
631 195000 8 0 15 0
622 176000 10 1 15 2
658 172000 10 1 15 0
628 84000 9 1 15 0
653 95000 10 0 15 1
640 123000 8 1 15 2
644 127000 10 1 15 1
645 171000 10 0 15 0
639 184000 7 1 15 2
620 108000 10 0 15 1
612 157000 9 1 14 1
619 158000 7 0 13 0
618 121000 7 0 13 2
617 136000 8 0 13 0
656 84000 10 1 13 2
630 200000 10 0 13 0
650 195000 10 1 12 2
651 132000 9 1 12 2
659 167000 9 0 12 2
621 166000 9 1 12 2
4 100000 6 0 12 3
627 146000 10 0 12 1
616 96000 10 1 11 2
6 100000 5 0 11 2
613 94000 9 1 11 0
624 178000 7 0 11 0
637 173000 9 1 11 0
655 99000 10 1 10 0
615 131000 9 1 10 1
633 163000 7 1 10 1
626 156000 9 1 10 2
641 144000 10 0 10 2
1 50000 3 0 10 9
647 183000 9 0 9 2
654 172000 10 1 9 1
642 97000 10 1 9 2
657 145000 8 1 9 1

Observations¶

  • Both the outliers column values look continuous and which might be valid/useful information for business analysis. Also this values can form a separate cluster/segment too.

  • So, we are not going to treat outliers in this case.

Standardization of features¶

What is feature scaling?¶

Feature scaling is a class of statistical techniques that, as the name implies, scales the features of our data so that they all have a similar range. You'll understand better if we look at an example:

If you have multiple independent variables like age, salary, and height, With their range as (18–100 Years), (25,000–75,000 Euros), and (1–2 Meters) respectively, feature scaling would help them all to be in the same range.

Why feature scaling is improtant in Unsupervised Learning?¶

Feature scaling is specially relevant in machine learning models that compute some sort of distance metric, like most clustering methods like K-Means.

So, scaling should be done to avoid the problem of one feature dominating over others because the unsupervised learning algorithm uses distance to find the similarity between data points.

Let's scale the data.¶

Standard Scaler¶

StandardScaler standardizes a feature by subtracting the mean and then scaling to unit variance. Unit variance means dividing all the values by the standard deviation.

SC.png

  1. Data standardization is the process of rescaling the attributes so that they have a mean of 0 and a variance of 1.
  2. The ultimate goal to perform standardization is to bring down all the features to a common scale without distorting the differences in the range of the values.
  3. In sklearn.preprocessing.StandardScaler(), centering and scaling happen independently on each feature.
In [37]:
from scipy.stats import zscore
subset_scaled_df=subset_df.apply(zscore)
subset_scaled_df.head(10)
Out[37]:
Avg_Credit_Limit Total_Credit_Cards Total_visits_bank Total_visits_online Total_calls_made
0 1.7402 -1.2492 -0.8605 -0.5475 -1.2515
1 0.4103 -0.7876 -1.4737 2.5205 1.8919
2 0.4103 1.0590 -0.8605 0.1343 0.1455
3 -0.1217 0.1357 -0.8605 -0.5475 0.1455
4 1.7402 0.5973 -1.4737 3.2023 -0.2037
5 -0.3876 -0.7876 -1.4737 -0.5475 1.5426
6 1.7402 0.1357 -1.4737 2.8614 -0.5530
7 -0.5206 -0.7876 -1.4737 -0.5475 -0.9023
8 -0.7866 -1.2492 -1.4737 -0.2066 -0.5530
9 -0.8398 -0.3259 -1.4737 -0.5475 1.1933
In [38]:
sc = StandardScaler()
subset_scaled_df1 = pd.DataFrame(sc.fit_transform(subset_df),columns=subset_df.columns)
subset_scaled_df1.head(10)
Out[38]:
Avg_Credit_Limit Total_Credit_Cards Total_visits_bank Total_visits_online Total_calls_made
0 1.7402 -1.2492 -0.8605 -0.5475 -1.2515
1 0.4103 -0.7876 -1.4737 2.5205 1.8919
2 0.4103 1.0590 -0.8605 0.1343 0.1455
3 -0.1217 0.1357 -0.8605 -0.5475 0.1455
4 1.7402 0.5973 -1.4737 3.2023 -0.2037
5 -0.3876 -0.7876 -1.4737 -0.5475 1.5426
6 1.7402 0.1357 -1.4737 2.8614 -0.5530
7 -0.5206 -0.7876 -1.4737 -0.5475 -0.9023
8 -0.7866 -1.2492 -1.4737 -0.2066 -0.5530
9 -0.8398 -0.3259 -1.4737 -0.5475 1.1933
  • zscore or StandardScaler() will give the same results. So, any method can be used for standardizing the features.

Unsupervised Learning Model Building for Clustering¶

K-Means Clustering¶

In [39]:
clusters = range(1, 10)
meanDistortions = []  # Create a empty list
errors = [] #Create empty list for errors 

for k in clusters:
    model = KMeans(n_clusters=k)  # Initialize KMeans
    model.fit(subset_scaled_df)  # Fit kMeans on the data
    prediction = model.predict(subset_df)  # Predict the model on the data
    distortion = (
        sum(np.min(cdist(subset_scaled_df, model.cluster_centers_, "euclidean"), axis=1))
        / subset_df.shape[0]  # Find distortion
    )

    meanDistortions.append(
        distortion
    )  # Append distortion values to the empty list created above
    
    errors.append(model.inertia_)

    print("Number of Clusters:", k, "\tAverage Distortion:", distortion)

print("\nCluster Error:", errors)

plt.plot(clusters, meanDistortions, "bo-")
plt.xlabel("k")  # Title of X-axis
plt.ylabel("Average Distortion")  # Title of y-axis
plt.title("Selecting k with the Elbow Method", fontsize=20)  # Title of the plot

Number of Clusters: 1 	Average Distortion: 2.0069222262503614
Number of Clusters: 2 	Average Distortion: 1.4571553548514269
Number of Clusters: 3 	Average Distortion: 1.1466276549150365
Number of Clusters: 4 	Average Distortion: 1.0463825294774465
Number of Clusters: 5 	Average Distortion: 0.9908683849620168
Number of Clusters: 6 	Average Distortion: 0.9430843103448057
Number of Clusters: 7 	Average Distortion: 0.9094700983137036
Number of Clusters: 8 	Average Distortion: 0.8922456881128548
Number of Clusters: 9 	Average Distortion: 0.8633274546350456

Cluster Error: [3300.0000000000005, 2040.9898164784952, 933.0437490000534, 780.7736895551768, 704.4759188657506, 642.1490713311521, 600.049543334103, 567.2035125446861, 534.2648682537475]
Out[39]:
Text(0.5, 1.0, 'Selecting k with the Elbow Method')

Observation¶

  • From the above Elbow Method graph, the possible number of clusters can be 3 or 4.
  • At k=3 the average distortion is 1.14 and at K=4 we have 1.04.
  • But the K value can be decided after checking Silhouette score
In [40]:
kev = KElbowVisualizer(
    estimator = KMeans(),
    k = 10,
    metric = "silhouette"
    # optional metric : "distortion"(default, "calinski_harabasz"
).fit(subset_scaled_df)

kev.show();

What is silhouette score?¶

Silhouette score is one of the methods for evaluating the quality of clusters created using clustering algorithms such as K-Means. The silhouette score is a measure of how similar an object is to its own cluster (cohesion) compared to other clusters (separation). Silhouette score has a range of [-1, 1].

  • Silhouette coefficients near +1 indicate that the sample is far away from the neighboring clusters.
  • Silhouette score near -1 indicate that those samples might have been assigned to the wrong cluster.

From the silhouette scores, it seems that 7 is a good value of k.Let's do deeper investigation for different K values using SilhouetteVisualizer

In [41]:
sil_score = []  # Create empty list
cluster_list = list(range(2, 10))  # Creating a list of range from 2 to 10
for n_clusters in cluster_list:
    clusterer = KMeans(n_clusters=n_clusters)  # Initializing KMeans algorithm
    preds = clusterer.fit_predict((subset_scaled_df))  # Predicting on the data
    # centers = clusterer.cluster_centers_
    score = silhouette_score(subset_scaled_df, preds)  # Cacalculating silhouette score
    sil_score.append(score)  # Appending silhouette score to empty list created above
    print("For n_clusters = {}, silhouette score is {}".format(n_clusters, score))

plt.plot(cluster_list, sil_score)

For n_clusters = 2, silhouette score is 0.41842496663215445
For n_clusters = 3, silhouette score is 0.5157182558881063
For n_clusters = 4, silhouette score is 0.3556670619372605
For n_clusters = 5, silhouette score is 0.2726898791817692
For n_clusters = 6, silhouette score is 0.25583657571102003
For n_clusters = 7, silhouette score is 0.24847787091758058
For n_clusters = 8, silhouette score is 0.2266798472313902
For n_clusters = 9, silhouette score is 0.21852430691339222
Out[41]:
[<matplotlib.lines.Line2D at 0x2448f5a1820>]

How to interpret the results of SilhouetteVisualizer?¶

Below are the points that every cluster needs to satisfy for selecting the optimal K-value.

  • There should be no negative silhouette coefficent value
  • All the clusters should cross the avg silhouette score
  • All the clusters should have the different variations (width)
  • All the clusters should have different silhouette score

Note

  1. Width of each cluster refers to the size of each cluster. (Number of data points present in each cluster)
  2. Horizontal line refers to the Silhouette score of each cluster.
  3. Dotted red line is an Average Silhouette score.
In [42]:
for ii in range(9, 1, -1):
    visualizer = SilhouetteVisualizer(KMeans(ii, random_state=1))
    visualizer.fit(subset_scaled_df)
    visualizer.show()

Observation

  • From the above plot, we can observe that, all the cluster values are more than the average Silhouette score
  • But some clusters are very thin in many cluster value centers.
  • If at all the cluster visualization looks good, but no elbow at many K levels
  • So overall K=3 looks good in many ways like above average score, elbow has a knick and cluster size is perfect too.

Applying KMeans clustering for k=3¶

In [43]:
%%time

# Creating Kmeans model with 3 clusters and fitting our scaled data to the the model
kmeans = KMeans(n_clusters=3, random_state=0)
kmeans.fit(subset_scaled_df)

CPU times: total: 344 ms
Wall time: 102 ms
Out[43]:
KMeans(n_clusters=3, random_state=0)
In [44]:
kmeans.labels_
Out[44]:
array([0, 1, 0, 0, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2,
       2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
       2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2])
In [45]:
# adding kmeans cluster labels to the original dataframe
df["K_means_segments"] = kmeans.labels_
In [46]:
# adding kmeans cluster labels to the original dataframe
subset_df["K_means_segments"] = kmeans.labels_
In [47]:
K_Means_df = df
K_Means_df
Out[47]:
Customer Key Avg_Credit_Limit Total_Credit_Cards Total_visits_bank Total_visits_online Total_calls_made K_means_segments
0 87073 100000 2 1 1 0 0
1 38414 50000 3 0 10 9 1
2 17341 50000 7 1 3 4 0
3 40496 30000 5 1 1 4 0
4 47437 100000 6 0 12 3 2
... ... ... ... ... ... ... ...
655 51108 99000 10 1 10 0 2
656 60732 84000 10 1 13 2 2
657 53834 145000 8 1 9 1 2
658 80655 172000 10 1 15 0 2
659 80150 167000 9 0 12 2 2

660 rows × 7 columns

Cluster Profiling¶

In [48]:
df[df["K_means_segments"] == 0]
Out[48]:
Customer Key Avg_Credit_Limit Total_Credit_Cards Total_visits_bank Total_visits_online Total_calls_made K_means_segments
0 87073 100000 2 1 1 0 0
2 17341 50000 7 1 3 4 0
3 40496 30000 5 1 1 4 0
228 54838 10000 7 2 0 0 0
229 35254 8000 7 2 1 4 0
... ... ... ... ... ... ... ...
607 90191 34000 6 3 1 0 0
608 49341 47000 4 4 1 0 0
609 11562 38000 4 3 2 0 0
610 16253 39000 7 2 1 0 0
611 80623 46000 7 4 1 3 0

386 rows × 7 columns

In [49]:
cluster_profile = df.groupby("K_means_segments").mean()
In [50]:
cluster_profile["count_in_each_segment"] = (
    df.groupby("K_means_segments")["Customer Key"].count().values
)
In [51]:
cluster_profile["count_in_each_segment"] = (
    df.groupby("K_means_segments").count().values
)
In [52]:
# let's display cluster profiles
cluster_profile.style.highlight_max(color="lightgreen", axis=0)
Out[52]:
  Customer Key Avg_Credit_Limit Total_Credit_Cards Total_visits_bank Total_visits_online Total_calls_made count_in_each_segment
K_means_segments              
0 54881.329016 33782.383420 5.515544 3.489637 0.981865 2.000000 386
1 55239.830357 12174.107143 2.410714 0.933036 3.553571 6.870536 224
2 56708.760000 141040.000000 8.740000 0.600000 10.900000 1.080000 50

Observation¶

  • Cluster 0 has 386 customers
  • Cluster 1 has 224 customers
  • Cluster 2 has 50 customers
In [53]:
import plotly as py
import plotly.graph_objs as go

trace1 = go.Scatter3d(
    x= df['Total_visits_bank'],
    y= df['Total_visits_online'],
    z= df['Total_calls_made'],
    mode='markers',
     marker=dict(
        color = df['K_means_segments'], 
        size= 20,
        line=dict(
            color= df['K_means_segments'],
            width= 12
        ),
        opacity=0.8
     ),

)
d = [trace1]
layout = go.Layout(
    title= 'K-Means Clusters',
    scene = dict(
            xaxis = dict(title  = 'Bank Visits'),
            yaxis = dict(title  = 'Online Visits'),
            zaxis = dict(title  = 'Customer Service Calls')
        )
)
fig = go.Figure(data=d, layout=layout)
py.offline.iplot(fig)
In [54]:
fig, axes = plt.subplots(1, 5, figsize=(16, 6))
fig.suptitle("Boxplot of numerical variables for each cluster")
counter = 0
for ii in range(5):
    sns.boxplot(ax=axes[ii], y=df[num_cols[counter]], x=df["K_means_segments"])
    counter = counter + 1

fig.tight_layout(pad=2.0)

Insights of KMeans Clustering¶

  • Cluster 0:

  • This Cluster customers prefer to handle bank transactions in person rather than calling or doing it online.

  • Also they have the medium number of credit cards and available credit.
  • Average Credit Limit is between 33k with 75% of the customers in this cluster\segment have an average >= than 12.5k, and 25% of the customers have an average credit limit >= 50k.
  • Total_Credit_Cards range between 2 and 7 with 75% of the customers in this cluster\segment have at least 4 credit cards.
  • Total_visits_Bank range betwenn 1 and 5 with 75% of the customers in this cluster\segment visit the bank in person for at least 3 times.
  • Total_Visits_Online count range between 0 and 3 with 75% of the customers in this cluster\segment login <=2 times.
  • Total_Calls_Made count range between 0 and 4 with 75% of the customers in this cluster\segment calls the bank or customer service department <= 3times.
  • Since group 0 use the online banking the least, bank should promote online banking more to group 0 in order for them to use it.

  • Cluster 1:

  • This Cluster Customers prefer to call the bank to solve any issues rather than visiting the bank or online banking.

  • Also they have the low number of credit cards and low average credit.
  • Average Credit Limit is between 3k and about 20k with 75% of the customers in this cluster\segment have an average <= than 15k.
  • Total_Credit_Cards range between 1 and 4 with 75% of the customers in this cluster\segment have at most 3 credit cards..
  • Total_visits_Bank range between 0 and 2 and visiting frequency is little less compare to cluster 0.
  • Total_Visits_Online count range between 1 and 5 with 75% of the customers in this cluster\segment login <=4 times.

  • Total_Calls_Made count range between 1 and 10 with 75% of the customers in this cluster\segment calls the bank or its customer service department at least 5 times.

  • Cluster 2:

  • This Cluster Customers to do online/digital banking to solve their issues rather than calling or visiting the bank.

  • Also they have a high average credit limit and high number of cards.
  • Average Credit Limit is approximately between 80k and 200k with 75% of the customers in this cluster\segment have an average >= than 110k.
  • Total_Credit_Cards range between 5 and 10 with 75% of the customers in this cluster\segment have at least 8 credit cards.
  • Total_visits_Bank range betwenn 0 and 1, customers in this cluster rarely visit the bank in person and have the lowest in person bank visiting frequency.
  • Total_Visits_Online count range between 6 and 15 with 75% of the customers in this cluster\segment at least 8 times.
  • Total_Calls_Made count range between 0 and 3 with 75% of the customers in this cluster\segment calls the bank or its customer service department at most 2 times.

PCA - Principal Component Analysis¶

In [55]:
covMatrix = np.cov(subset_scaled_df,rowvar=False)
print(covMatrix)

[[ 1.00151745  0.60978441 -0.10046453  0.55222122 -0.41498065]
 [ 0.60978441  1.00151745  0.31627526  0.16801286 -0.65223974]
 [-0.10046453  0.31627526  1.00151745 -0.55269882 -0.50678371]
 [ 0.55222122  0.16801286 -0.55269882  1.00151745  0.12749255]
 [-0.41498065 -0.65223974 -0.50678371  0.12749255  1.00151745]]
In [56]:
pca = PCA(n_components=5)
pca.fit(subset_scaled_df)
Out[56]:
PCA(n_components=5)
In [57]:
#The eigen Values
print(pca.explained_variance_)

[2.29029906 1.87416377 0.3205531  0.27826563 0.24430569]
In [58]:
#The eigen Vectors
print(pca.components_)

[[ 0.48885901  0.59767943  0.28049227  0.11178284 -0.55912866]
 [ 0.40323959  0.03017103 -0.58658661  0.66516108  0.22352726]
 [ 0.00346147 -0.28498293 -0.6135219  -0.3049481  -0.67035086]
 [ 0.30861651 -0.74135216  0.44527763  0.31838769 -0.23560546]
 [-0.70933671  0.10512237  0.05058609  0.59220036 -0.36404738]]
In [59]:
#And the percentage of variation explained by each eigen Vector
print(pca.explained_variance_ratio_)

[0.45736578 0.37426483 0.06401348 0.0555688  0.04878711]
In [60]:
plt.bar(list(range(1,6)),pca.explained_variance_ratio_,alpha=0.5, align='center')
plt.ylabel('Variation explained')
plt.xlabel('eigen Value')
plt.show()
In [61]:
plt.step(list(range(1,6)),np.cumsum(pca.explained_variance_ratio_), where='mid')
plt.ylabel('Cum of variation explained')
plt.xlabel('eigen Value')
plt.show()

Dimensionality Reduction¶

Now 3 dimensions seems very reasonable. With 3 variables we can explain over 95% of the variation in the original data!

In [62]:
pca3 = PCA(n_components=3)
pca3.fit(subset_scaled_df)
print(pca3.components_)
print(pca3.explained_variance_ratio_)
Xpca3 = pca3.transform(subset_scaled_df)

[[ 0.48885901  0.59767943  0.28049227  0.11178284 -0.55912866]
 [ 0.40323959  0.03017103 -0.58658661  0.66516108  0.22352726]
 [ 0.00346147 -0.28498293 -0.6135219  -0.3049481  -0.67035086]]
[0.45736578 0.37426483 0.06401348]
In [63]:
PCA_df = pd.DataFrame(Xpca3)
PCA_df.head()
Out[63]:
0 1 2
0 0.5013 0.5248 1.8959
1 -1.4596 3.1056 -0.9068
2 0.5258 0.8240 0.0890
3 -0.3623 0.1281 0.5582
4 1.2662 3.6687 -0.1000
In [64]:
sns.pairplot(PCA_df)
Out[64]:
<seaborn.axisgrid.PairGrid at 0x24490c134f0>

Hierarchical Clustering¶

  • It is a Connectivity based clustering which creates clusters in a hierarchical tree like structure
  • Any type of distance measure can be used as a measure of similarity.
  • The Cluster tree like output is called Dendogram.

Let's find the Cophenetic correlation for different distances with different linkage methods.

What is Cophenetic correlation?¶

The cophenetic correlation coefficient is a correlation coefficient between the cophenetic distances(Dendrogramic distance) obtained from the tree, and the original distances used to construct the tree. It is a measure of how faithfully a dendrogram preserves the pairwise distances between the original unmodeled data points.

The cophenetic distance between two observations is represented in a dendrogram by the height of the link at which those two observations are first joined. That height is the distance between the two subclusters that are merged by that link.

Cophenetic correlation is the way to compare two or more dendrograms.

In [65]:
# list of distance metrics
distance_metrics = ["euclidean", "chebyshev", "mahalanobis", "cityblock"]

# list of linkage methods
linkage_methods = ["single", "complete", "average", "weighted"]

high_cophenet_corr = 0
high_dm_lm = [0, 0]

for dm in distance_metrics:
    for lm in linkage_methods:
        Z = linkage(subset_scaled_df, metric=dm, method=lm)
        c, coph_dists = cophenet(Z, pdist(subset_scaled_df))
        print(
            "Cophenetic correlation for {} distance and {} linkage is {}.".format(
                dm.capitalize(), lm, c
            )
        )
        if high_cophenet_corr < c:
            high_cophenet_corr = c
            high_dm_lm[0] = dm
            high_dm_lm[1] = lm

Cophenetic correlation for Euclidean distance and single linkage is 0.7391220243806552.
Cophenetic correlation for Euclidean distance and complete linkage is 0.8599730607972423.
Cophenetic correlation for Euclidean distance and average linkage is 0.8977080867389372.
Cophenetic correlation for Euclidean distance and weighted linkage is 0.8861746814895477.
Cophenetic correlation for Chebyshev distance and single linkage is 0.7382354769296767.
Cophenetic correlation for Chebyshev distance and complete linkage is 0.8533474836336782.
Cophenetic correlation for Chebyshev distance and average linkage is 0.8974159511838106.
Cophenetic correlation for Chebyshev distance and weighted linkage is 0.8913624010768603.
Cophenetic correlation for Mahalanobis distance and single linkage is 0.7058064784553606.
Cophenetic correlation for Mahalanobis distance and complete linkage is 0.5422791209801747.
Cophenetic correlation for Mahalanobis distance and average linkage is 0.8326994115042134.
Cophenetic correlation for Mahalanobis distance and weighted linkage is 0.7805990615142516.
Cophenetic correlation for Cityblock distance and single linkage is 0.7252379350252723.
Cophenetic correlation for Cityblock distance and complete linkage is 0.8731477899179829.
Cophenetic correlation for Cityblock distance and average linkage is 0.896329431104133.
Cophenetic correlation for Cityblock distance and weighted linkage is 0.8825520731498188.
In [66]:
# printing the combination of distance metric and linkage method with the highest cophenetic correlation
print(
    "Highest cophenetic correlation is {}, which is obtained with {} distance and {} linkage.".format(
        high_cophenet_corr, high_dm_lm[0].capitalize(), high_dm_lm[1]
    )
)

Highest cophenetic correlation is 0.8977080867389372, which is obtained with Euclidean distance and average linkage.

Let's explore different linkage methods with Euclidean distance only.

In [67]:
# list of linkage methods
linkage_methods = ["single", "complete", "average", "centroid", "ward", "weighted"]

high_cophenet_corr = 0
high_dm_lm = [0, 0]

for lm in linkage_methods:
    Z = linkage(subset_scaled_df, metric="euclidean", method=lm)
    c, coph_dists = cophenet(Z, pdist(subset_scaled_df))
    print("Cophenetic correlation for {} linkage is {}.".format(lm, c))
    if high_cophenet_corr < c:
        high_cophenet_corr = c
        high_dm_lm[0] = "euclidean"
        high_dm_lm[1] = lm

Cophenetic correlation for single linkage is 0.7391220243806552.
Cophenetic correlation for complete linkage is 0.8599730607972423.
Cophenetic correlation for average linkage is 0.8977080867389372.
Cophenetic correlation for centroid linkage is 0.8939385846326323.
Cophenetic correlation for ward linkage is 0.7415156284827493.
Cophenetic correlation for weighted linkage is 0.8861746814895477.
In [68]:
# printing the combination of distance metric and linkage method with the highest cophenetic correlation
print(
    "Highest cophenetic correlation is {}, which is obtained with {} linkage.".format(
        high_cophenet_corr, high_dm_lm[1]
    )
)

Highest cophenetic correlation is 0.8977080867389372, which is obtained with average linkage.

We see that the cophenetic correlation is maximum with Euclidean distance and average linkage.

Let's see the dendrograms for the different linkage methods.

In [69]:
# list of linkage methods
linkage_methods = ["single", "complete", "average", "centroid", "ward", "weighted"]

# lists to save results of cophenetic correlation calculation
compare_cols = ["Linkage", "Cophenetic Coefficient"]

# to create a subplot image
fig, axs = plt.subplots(len(linkage_methods), 1, figsize=(15, 30))

# We will enumerate through the list of linkage methods above
# For each linkage method, we will plot the dendrogram and calculate the cophenetic correlation
for i, method in enumerate(linkage_methods):
    Z = linkage(subset_scaled_df, metric="euclidean", method=method)

    dendrogram(Z, ax=axs[i])
    axs[i].set_title(f"Dendrogram ({method.capitalize()} Linkage)")

    coph_corr, coph_dist = cophenet(Z, pdist(subset_scaled_df))
    axs[i].annotate(
        f"Cophenetic\nCorrelation\n{coph_corr:0.2f}",
        (0.80, 0.80),
        xycoords="axes fraction",
    )

Observations

  • The cophenetic correlation is highest for average and centroid, weighted linkage methods have almost equal correlation.
  • 3 clusters appears to be the appropriate number of clusters from the dendrogram for Average linkage.
  • So, We will move ahead with average linkage and 3 clusters.

Observations

  • It looks like Ward linkage has more distinct and separated clusters.
  • The cophenetic correlation is highest for average linkage method.
  • We will move ahead with average linkage for now and see how well the clusters were profiled.
  • Let's check the best number of clusters for this method from the dendogram. The longest vertical distance without any horizontal line passing through it is represented by the left blue leg, therefore this leg\line is selected and a horizontal line is drawn through it. The number of vertical lines this newly created horizontal line passes is equal to number of clusters.

The optimal number of clusters from a dendrogram can be obtained by deciding where to cut the cluster tree. Generally, the cluster tree is cut where dendrogram height is maximum as it generally corresponds to distinct and homogeneous clusters. The dendrogram for average linkage had the highest cophenetic correlation. So we can use average linkage and have chosen 3 clusters as the dendrogram height is pretty high. (The maximum height is for 2, but that would not be meaningful as it would give one clusters with too many data and its hard to contact customers as per their preferences).

Lets visualize the dendrogram cut for average link in the below plot

In [91]:
# list of linkage methods
linkage_methods = ["average"]

# lists to save results of cophenetic correlation calculation
compare_cols = ["Linkage", "Cophenetic Coefficient"]

# to create a subplot image
fig, axs = plt.subplots(len(linkage_methods), 1, figsize=(15, 15))

# We will enumerate through the list of linkage methods above
# For each linkage method, we will plot the dendrogram and calculate the cophenetic correlation
for i, method in enumerate(linkage_methods):
    Z = linkage(subset_scaled_df, metric="euclidean", method=method)

    dendrogram(Z, ax=axs)
    axs.set_title(f"Dendrogram ({method.capitalize()} Linkage)")

    coph_corr, coph_dist = cophenet(Z, pdist(subset_scaled_df))
    axs.annotate(
        f"Cophenetic\nCorrelation\n{coph_corr:0.2f}",
        (0.80, 0.80),
        xycoords="axes fraction",
    )
dendrogram(Z, color_threshold=3.2)
plt.axhline(y=3.2, c="red", lw=1, linestyle="dashdot")
Out[91]:
<matplotlib.lines.Line2D at 0x24497dc1e80>

Hierarchical Clustering¶

In [71]:
HCmodel = AgglomerativeClustering(n_clusters=3, affinity="euclidean", linkage="average")
HCmodel.fit(subset_scaled_df)
Out[71]:
AgglomerativeClustering(linkage='average', n_clusters=3)
In [72]:
subset_scaled_df["HC_Clusters"] = HCmodel.labels_
df["HC_Clusters"] = HCmodel.labels_
In [73]:
HC_Model_df = df
HC_Model_df
Out[73]:
Customer Key Avg_Credit_Limit Total_Credit_Cards Total_visits_bank Total_visits_online Total_calls_made K_means_segments HC_Clusters
0 87073 100000 2 1 1 0 0 0
1 38414 50000 3 0 10 9 1 2
2 17341 50000 7 1 3 4 0 0
3 40496 30000 5 1 1 4 0 0
4 47437 100000 6 0 12 3 2 1
... ... ... ... ... ... ... ... ...
655 51108 99000 10 1 10 0 2 1
656 60732 84000 10 1 13 2 2 1
657 53834 145000 8 1 9 1 2 1
658 80655 172000 10 1 15 0 2 1
659 80150 167000 9 0 12 2 2 1

660 rows × 8 columns

Cluster Profiling¶

In [74]:
cluster_profile = df.groupby("HC_Clusters").mean()
In [75]:
cluster_profile["count_in_each_segments"] = (
    df.groupby("HC_Clusters")["Customer Key"].count().values
)
In [76]:
df['HC_Clusters'].value_counts()
Out[76]:
0    387
2    223
1     50
Name: HC_Clusters, dtype: int64

Average linkage seems have a good cluster, but ward always give homogeneous clusters. So we can just try with ward.

Lets visualize the dendrogram cut for ward linkage in the below plot

In [77]:
# list of linkage methods
linkage_methods = ["ward"]

# lists to save results of cophenetic correlation calculation
compare_cols = ["Linkage", "Cophenetic Coefficient"]

# to create a subplot image
fig, axs = plt.subplots(len(linkage_methods), 1, figsize=(15, 15))

# We will enumerate through the list of linkage methods above
# For each linkage method, we will plot the dendrogram and calculate the cophenetic correlation
for i, method in enumerate(linkage_methods):
    Z = linkage(subset_scaled_df, metric="euclidean", method=method)

    dendrogram(Z, ax=axs)
    axs.set_title(f"Dendrogram ({method.capitalize()} Linkage)")

    coph_corr, coph_dist = cophenet(Z, pdist(subset_scaled_df))
    axs.annotate(
        f"Cophenetic\nCorrelation\n{coph_corr:0.2f}",
        (0.80, 0.80),
        xycoords="axes fraction",
    )
dendrogram(Z, color_threshold=18)
plt.axhline(y=18, c="red", lw=1, linestyle="dashdot")
Out[77]:
<matplotlib.lines.Line2D at 0x2449cac2610>
In [92]:
%%time 
# Agglomerative Clustering with 3 clusters, ward linkage and euclidean as the metric 
HCmodel = AgglomerativeClustering(n_clusters=3, affinity="euclidean", linkage="ward")
HCmodel.fit(subset_scaled_df)

# Calculting the silhouette score for the Agglomerative clustering
print(silhouette_score(subset_scaled_df, HCmodel.labels_, metric="euclidean"))

0.5413025796835845
CPU times: total: 78.1 ms
Wall time: 52.6 ms
In [93]:
subset_scaled_df["HC_Clusters"] = HCmodel.labels_
df["HC_Clusters"] = HCmodel.labels_
In [94]:
df['HC_Clusters'].value_counts()
Out[94]:
0    387
1    223
2     50
Name: HC_Clusters, dtype: int64

Cluster Profiling¶

In [95]:
cluster_profile = df.groupby("HC_Clusters").mean()
In [96]:
cluster_profile["count_in_each_segments"] = (
    df.groupby("HC_Clusters")["Customer Key"].count().values
)
In [97]:
# let's see the names of the countries in each cluster
for cl in df["HC_Clusters"].unique():
    print(
        "The",
        df[df["HC_Clusters"] == cl]["Customer Key"].nunique(),
        "countries in cluster",
        cl,
        "are:",
    )
    print(df[df["HC_Clusters"] == cl]["Customer Key"].unique())
    print("-" * 100, "\n")

The 385 countries in cluster 0 are:
[87073 17341 40496 54838 35254 46635 97825 83125 35483 15129 83290 56486
 31903 45909 14263 46813 81878 35549 85799 39122 81531 69965 18595 44398
 32352 40898 27101 33457 45088 23302 27408 65372 21531 56843 17165 89328
 20072 71402 47496 24808 17036 67193 34423 97109 55382 51811 53936 66504
 53207 18514 51319 36340 36934 95925 49771 22919 21233 74544 52025 45652
 73952 49418 77026 49331 75775 54906 94666 11698 34677 95610 41380 38033
 85337 38994 67911 92956 77641 57565 53814 30712 19785 31384 16374 50878
 78002 83459 91987 51552 72156 24998 45673 11596 87485 28414 81863 33240
 11466 23881 44645 49844 92782 22824 26767 26678 50412 17933 34495 47437
 22610 41159 64672 62483 85614 96548 19137 69028 70779 38244 67046 64897
 46223 36628 17565 77381 11799 81940 66706 87838 94437 33790 44402 29886
 66804 47866 61996 15318 89635 71681 71862 96186 22348 36243 88807 82376
 98126 80347 17649 62807 92522 57459 44579 45476 61994 11398 24702 27824
 45878 72431 19215 23409 16418 85122 55060 55478 65574 31113 96929 78912
 68439 62864 31515 77954 88207 78618 31551 75792 29864 45440 97954 90189
 55090 17703 33991 88884 45808 50706 92140 88123 53932 65908 25321 87456
 48602 97530 48657 76209 49913 53002 61122 82807 93496 64519 31950 23110
 96297 28408 37252 41287 52460 26604 58019 87219 36839 12663 48667 42887
 14439 60851 41266 37438 65747 81166 20570 14816 11265 24980 37934 70707
 84351 89446 17325 64774 53166 45341 94595 55170 92489 92933 36504 40508
 15798 70101 77613 84360 48402 46776 67258 44804 29919 65781 12456 62649
 74446 36632 76024 75065 51682 18397 29102 56367 95147 44379 76957 42921
 23102 61324 49690 20043 44144 53552 62530 41741 22842 65825 77826 61216
 83192 82023 73000 64550 90131 17382 27117 94529 21717 81910 76492 43000
 48692 27476 15086 43034 99131 13140 99437 91242 39285 63710 90860 35585
 58708 57451 69868 43679 30256 26334 47848 17377 39644 29176 55706 51771
 83585 51867 68040 75417 34775 85645 83545 44157 38125 75398 90999 70376
 33295 80942 26493 97850 43841 79885 59316 83466 81510 35268 11734 88411
 96269 87683 26063 42479 58116 67282 84888 75366 14377 59074 96534 31870
 24748 68920 67637 60839 59170 90586 56270 87670 47703 35421 58511 76398
 93310 36836 46373 94700 67860 99473 68862 93381 46548 74083 48660 13720
 72339 99284 47198 67415 44403 58276 85234 31948 90191 49341 11562 16253
 80623]
---------------------------------------------------------------------------------------------------- 

The 223 countries in cluster 1 are:
[38414 58634 37376 82490 44770 52741 52326 92503 25084 68517 55196 62617
 96463 39137 14309 29794 87241 30507 61061 24001 68067 65034 14854 81130
 29112 13999 32550 82164 61517 28254 30888 46388 74126 52142 37659 83132
 20119 52363 50769 68502 99026 83326 62040 37252 74625 51182 60301 96386
 43886 78503 68419 32828 17937 71632 81566 29759 36929 70248 91673 61355
 60403 85868 76205 66524 69214 21976 35149 27120 18821 33187 93482 90168
 71881 59656 12026 99589 38970 57990 39447 79694 79403 47296 37559 38165
 49198 18007 59619 37016 91099 74704 25742 11937 52736 88338 18916 92501
 96213 26599 73007 97935 26089 14946 74795 73435 41634 84069 83244 87291
 18086 33369 15310 98499 35256 89007 93997 16577 25440 81116 63663 69811
 36111 39454 70199 11602 49697 28701 61627 34103 14248 31256 45583 52750
 95507 23743 53410 53898 66200 58389 61347 59151 37802 60475 95489 77758
 23768 87471 85707 97951 54785 97011 35103 18564 61009 24054 63751 52758
 78473 80457 59783 64241 32374 97536 33110 36978 54281 98602 97687 28842
 38410 38261 20524 37671 25330 41787 11412 55892 95495 41946 86410 76718
 98969 77143 38205 53851 52783 63405 48510 97463 18145 14398 98288 69704
 29058 15546 16715 87350 13215 20593 56624 33317 99596 72430 16676 40486
 90958 67212 44226 94251 61776 55275 18609 54477 12122 28208 68003 79632
 73811 72892 51773 96163 61234 55849 56156]
---------------------------------------------------------------------------------------------------- 

The 50 countries in cluster 2 are:
[47437 48370 94391 50598 40019 77910 89832 98216 54495 47650 32107 84192
 53916 32584 97285 20337 15585 20620 75009 76203 33837 14916 97935 16180
 49493 70974 40217 88442 17538 90839 99843 27212 91575 60190 18519 48762
 58392 79953 13315 30570 78996 78404 28525 51826 65750 51108 60732 53834
 80655 80150]
----------------------------------------------------------------------------------------------------

Both linkage method average & ward have almost same type of clusters. So we can use anything.

In [98]:
# lets display cluster profile
cluster_profile.style.highlight_max(color="lightgreen", axis=0)
Out[98]:
  Customer Key Avg_Credit_Limit Total_Credit_Cards Total_visits_bank Total_visits_online Total_calls_made K_means_segments count_in_each_segments
HC_Clusters                
0 54925.966408 33713.178295 5.511628 3.485788 0.984496 2.005168 0.002584 387
1 55163.973094 12197.309417 2.403587 0.928251 3.560538 6.883408 1.000000 223
2 56708.760000 141040.000000 8.740000 0.600000 10.900000 1.080000 2.000000 50
In [99]:
sns.pairplot(subset_scaled_df, diag_kind='kde', hue='HC_Clusters');
In [100]:
fig, axes = plt.subplots(1, 5, figsize=(16, 6))
fig.suptitle("Boxplot of original numerical variables for each cluster")
counter = 0
for ii in range(5):
    sns.boxplot(
        ax=axes[ii],
        y=df[num_cols[counter]],
        x=df["HC_Clusters"],
    )
    counter = counter + 1

fig.tight_layout(pad=2.0)

Insights of Hierarchical Clustering¶

  • Cluster 0:

  • This Cluster customers prefer to handle bank transactions in person rather than calling or doing it online.

  • Also they have the medium number of credit cards and available credit.
  • Average Credit Limit is between 33k with 75% of the customers in this cluster\segment have an average >= than 12.5k, and 25% of the customers have an average credit limit >= 50k.
  • Total_Credit_Cards range between 2 and 7 with 75% of the customers in this cluster\segment have at least 4 credit cards.
  • Total_visits_Bank range between 1 and 5 with 75% of the customers in this cluster\segment visit the bank in person for at least 3 times.
  • Total_Visits_Online count range between 0 and 3 with 75% of the customers in this cluster\segment login <=2 times.
  • Total_Calls_Made count range between 0 and 4 with 75% of the customers in this cluster\segment calls the bank or customer service department <= 3times.
  • Since group 0 use the online banking the least, bank should promote online banking more to group 0 in order for them to use it.

  • Cluster 1:

  • This Cluster Customers prefer to call the bank to solve any issues rather than visiting the bank or online banking.

  • Also they have the low number of credit cards and low average credit.
  • Average Credit Limit is between 3k and about 20k with 75% of the customers in this cluster\segment have an average <= than 15k.
  • Total_Credit_Cards range between 1 and 4 with 75% of the customers in this cluster\segment have at most 3 credit cards..
  • Total_visits_Bank range between 0 and 2 and visiting frequency is little less compare to cluster 0.
  • Total_Visits_Online count range between 1 and 5 with 75% of the customers in this cluster\segment login <=4 times.

  • Total_Calls_Made count range between 1 and 10 with 75% of the customers in this cluster\segment calls the bank or its customer service department at least 5 times.

  • Cluster 2:

  • This Cluster Customers do online/digital banking to solve their issues rather than calling or visiting the bank.

  • Also they have a high average credit limit and high number of cards.
  • Average Credit Limit is approximately between 80k and 200k with 75% of the customers in this cluster\segment have an average >= than 110k.
  • Total_Credit_Cards range between 5 and 10 with 75% of the customers in this cluster\segment have at least 8 credit cards.
  • Total_visits_Bank range betwenn 0 and 1, customers in this cluster rarely visit the bank in person and have the lowest in person bank visiting frequency.
  • Total_Visits_Online count range between 6 and 15 with 75% of the customers in this cluster\segment at least 8 times.
  • Total_Calls_Made count range between 0 and 3 with 75% of the customers in this cluster\segment calls the bank or its customer service department at most 2 times.

Additional Analysis¶

In [101]:
PCA_Cluster_df = PCA_df.copy()
PCA_Cluster_df['PCA_HC_Clusters'] = HCmodel.labels_

#Renaming the columns for easy identification
PCA_Cluster_df = PCA_Cluster_df.rename(columns={0:'component_1', 1:'component_2', 2:'component_3'})

PCA_Cluster_df
Out[101]:
component_1 component_2 component_3 PCA_HC_Clusters
0 0.5013 0.5248 1.8959 0
1 -1.4596 3.1056 -0.9068 1
2 0.5258 0.8240 0.0890 0
3 -0.3623 0.1281 0.5582 0
4 1.2662 3.6687 -0.1000 2
... ... ... ... ...
655 3.0385 2.6662 -0.0923 2
656 2.5672 3.3417 -0.8738 2
657 2.8514 2.9831 0.0449 2
658 4.1783 4.5829 -0.6053 2
659 3.1604 4.3510 -0.2544 2

660 rows × 4 columns

In [102]:
sns.pairplot(PCA_Cluster_df, diag_kind='kde', hue='PCA_HC_Clusters');
In [103]:
import plotly as py
import plotly.graph_objs as go

trace1 = go.Scatter3d(
    x= PCA_Cluster_df['component_1'],
    y= PCA_Cluster_df['component_2'],
    z= PCA_Cluster_df['component_3'],
    mode='markers',
     marker=dict(
        color = PCA_Cluster_df['PCA_HC_Clusters'], 
        size= 20,
        line=dict(
            color= PCA_Cluster_df['PCA_HC_Clusters'],
            width= 12
        ),
        opacity=0.8
     ),

)
d = [trace1]
layout = go.Layout(
    title= 'PCA HC Clusters',
    scene = dict(
            xaxis = dict(title  = 'Component 1'),
            yaxis = dict(title  = 'Component 2'),
            zaxis = dict(title  = 'Component 3')
        )
)
fig = go.Figure(data=d, layout=layout)
py.offline.iplot(fig)
In [113]:
sns.scatterplot(
    data=PCA_Cluster_df,
    x="component_1",
    y="component_2",
    hue="PCA_HC_Clusters",
    palette="rainbow",
)
plt.legend(bbox_to_anchor=(1, 1))
Out[113]:
<matplotlib.legend.Legend at 0x2449fd1bb50>
In [114]:
PCA_Cluster_df.boxplot(by='PCA_HC_Clusters', layout=(3,2), figsize=(10,14));
In [115]:
subset_scaled_df.boxplot(by='HC_Clusters', layout=(3,2), figsize=(10,14));

Analyzing the duplicated Customer Key¶

In [116]:
df[df["Customer Key"].isin([47437, 37252, 97935, 96929, 50706])].sort_values('Customer Key')
Out[116]:
Customer Key Avg_Credit_Limit Total_Credit_Cards Total_visits_bank Total_visits_online Total_calls_made K_means_segments HC_Clusters
48 37252 6000 4 0 2 8 1 1
432 37252 59000 6 2 1 2 0 0
4 47437 100000 6 0 12 3 2 2
332 47437 17000 7 3 1 0 0 0
411 50706 44000 4 5 0 2 0 0
541 50706 60000 7 5 2 2 0 0
391 96929 13000 4 5 0 0 0 0
398 96929 67000 6 2 2 2 0 0
104 97935 17000 2 1 2 10 1 1
632 97935 187000 7 1 7 0 2 2

Observations

  • The above Customer key are duplicate records with different values in all the other features
  • But three customer 47437, 37252, 97935 are assigned to different clusters
  • So bank should try to fix it, otherwise they will be approached by different markerting teams
  • If they are really different customers, then the Customer Key needs to be fixed as its a unique value.

Comparison of KMeans and Hierarchical Clustering¶

In [117]:
fig, (ax1,ax2) = plt.subplots(1,2, figsize=(15,10))

ax1 = plt.subplot(1,2,1)
plt.title('KMeans Predicted Classes')
sns.scatterplot(x='Avg_Credit_Limit', y='Total_Credit_Cards', style='K_means_segments', hue='K_means_segments',data=K_Means_df,ax=ax1)

ax2 = plt.subplot(1,2,2)
plt.title('Hierarchical Predicted Classes')
sns.scatterplot(x='Avg_Credit_Limit', y='Total_Credit_Cards', style = 'HC_Clusters',hue='HC_Clusters', data=HC_Model_df, ax=ax2)

plt.show()
In [118]:
fig, (ax1,ax2) = plt.subplots(1,2, figsize=(15,10))

ax1 = plt.subplot(1,2,1)
plt.title('KMeans Predicted Classes')
sns.scatterplot(x='Avg_Credit_Limit', y='Total_visits_online', style='K_means_segments', hue='K_means_segments',data=K_Means_df,ax=ax1)

ax2 = plt.subplot(1,2,2)
plt.title('Hierarchical Predicted Classes')
sns.scatterplot(x='Avg_Credit_Limit', y='Total_visits_online', style = 'HC_Clusters',hue='HC_Clusters', data=HC_Model_df, ax=ax2)

plt.show()
In [119]:
fig, (ax1,ax2) = plt.subplots(1,2, figsize=(15,10))

ax1 = plt.subplot(1,2,1)
plt.title('KMeans Predicted Classes')
sns.scatterplot(x='Total_visits_bank', y='Total_calls_made', style='K_means_segments', hue='K_means_segments',data=K_Means_df,ax=ax1)

ax2 = plt.subplot(1,2,2)
plt.title('Hierarchical Predicted Classes')
sns.scatterplot(x='Total_visits_bank', y='Total_calls_made', style = 'HC_Clusters',hue='HC_Clusters', data=HC_Model_df, ax=ax2)

plt.show()

Comparing K-means and Agglomerative clustering¶

  • K Means clustering needed advance knowledge of K i.e. no. of clusters one want to divide your data.In hierarchical clustering one can stop at any number of clusters, one find appropriate by interpreting the dendrogram.

  • K-Means Clustering is Centroid based, Hierarchical Clustering is partition-based and can either be divisive (Top-down) or agglomerative(bottom-up).

  • K-Means Clustering - The Elbow method used to find clusters Hierarchical Clustering - Dendrogram is used

  • But The results of both of our clustering are almost similar for this dataset.

  • It might be possible to see a difference, when we have a very large dataset and that shape of clusters may differ a little.

Time for execution:

  • K-means took 102ms and Agglomerative clustering took 53ms.

Distinct clusters:

  • Both clustering technique KMeans and Agglomerative gave same clusters.

Silhouette scores:

  • K-means clustering had a better silhoutte score with 0.51 compared to the 0.54 for the agglomerative clustering.

Number of observations:

There are almost similar clusters in both algorithms. Below are the counts.

  • Cluster 0 has 386 customers in KMeans and 387 in Agglomerative.
  • Cluster 1 has 224 customers in KMeans and 223 in Agglomerative.
  • Cluster 2 has 50 customers in KMeans and 50 in Agglomerative.

Business recommendations and insights¶

  • AllLife bank can use any clustering technique KMeans or Hierarchical in segmenting its customers base, since both are equally optimal with respect to efficient(time wise), distinct clusters, and silhouette score.

  • As we see there are three categories/segments of customers with each segment having a preference for communication channel with bank, it is recommended that products are marketed to specific segment of customers through their preferred channel.

  • Also, additional services can be provided based on how they connect with bank and also based on their spending pattern which can deduced from average credit limit.

  • At the same time the bank should try to increase the credit limit and the number of credit cards for customers in these cluters perhaps by offering incentives, bonuses, or waiving some fees.

  • This approach should be more intensive to customers in cluster 0 since customers in cluster 0 have the least credit limit and the least number of cards..

  • Customers with high credit limit tend to vist online. Hence they can be targeted for online campaigns, coupons and accordingly products and services can be offered to them

  • Whereas customers with comparatively low credit limit make visits to bank more often, hence they can either be prompted with benefits of online banking or can be catered with in-bank offers/services and flyers

  • Customers with low credit limits have less frequency on online platform, they can be marketed with benefits of online banking and/or make customer call center reps aware of promotions and offers so that they can target this segment of customers

  • Based on how bank wants to promote its products and services, a segment of customers can be targeted as we know their prefered mode of commmunication with bank